As the rotor circulates in the launch loop, it ascends from ground level to the 100 kilometer altitude of the launch path, then descends back down to the surface for the ground anchored ambits. If the 3 kilogram-per-meter rotor moves at 14 km/second at altitude, it will gain energy, speed up, and stretch out as it descends. The stretch is because the rotor "flux" passing any given point must be 3 kg/m * 14 km/s = 42 tonnes per second.
How much does it speed up? If the surface is at 6378 km, and the launch path is at 6478 km (100 km altitude), then the energy difference per kilogram is ( 398600.44 km³/s² × ( 1/6378 km - 1/6478 km ) = 0.96474 km²/s² , which is added to the ½ × (14 km/s)² = 98 km²/s² (per kg) of the rotor at altitude, resulting in a ground level kinetic energy of 98.96474 km²/s² (per kg), corresponding to a rotor speed of 14.069 km/s, 69 m/s faster
- Note: A second order wrinkle is the rotation velocity of the Earth, 0.4724 km/s at 100 km and 0.4651 km/s at the surface (a speed difference of , which is added to the westbound rotor speed and subtracted from the eastbound rotor speed. That adds another 3 m/s to the westbound rotor at the surface.
The eastbound rotor has the same kinetic energy at altitude as the westbound rotor, and the same velocity with respect to a nonrotating inertial frame, but the rotating earth ...
MoreLater . If the easttbound rotor moves at a ground relative speed of 14 km/s, it is moving at 14.474 km/s in the inertial frame, and this is what provides track support. The westbound rotor will also be moving at 14.474 km/s in the inertial frame, it is moving at 14.948 km/s in the ground-referenced rotating frame. I MUST THINK THIS THROUGH.